Question

这个很简单，没有任何典型的答案可以冒出来。

我正在使用的工具的性质决定了我们使用MongoDB来存储＆＃39;设置＆＃39;大约25种不同的工具。每个工具都有自己的设置架构，因此每个文档都不同，但它们全部存储在同一个集合中，并在json架构绘制的同一编辑页面上进行编辑。

不知道字典的架构，我很难弄清楚如何迭代和消毒数据，特别是删除密码。

鉴于以下字典，并且知道其他字典可能有不同的模式，我怎么能遍历字典中的每一个项目并创建一个副本，相同的除了与任何键==＆＃34 ;密码＆＃34;去掉？

所以：

{
  "_enabled": true,
  "instances": [
    {
      "isdefault": true,
      "name": "dev",
      "password": "abc123",
      "url": "http://dev.example.com",
      "user": "buffy"
    },
    {
      "isdefault": false,
      "name": "prod",
      "password": "xxxxx",
      "url": "http://prod.example.com",
      "user": "spike"
    },
    {
      "isdefault": false,
      "name": "qa",
      "password": "dasddf",
      "url": "http://prod.example.com",
      "user": "willow"
    }
  ],
  "label": "MyServers"
}

应该导致：

{
  "_enabled": true,
  "instances": [
    {
      "isdefault": true,
      "name": "dev",
      "url": "http://dev.example.com",
      "user": "buffy"
    },
    {
      "isdefault": false,
      "name": "prod",
      "url": "http://prod.example.com",
      "user": "spike"
    },
    {
      "isdefault": false,
      "name": "qa",
      "url": "http://prod.example.com",
      "user": "willow"
    }
  ],
  "label": "MyServers"
}

Answer 1

首先对dict进行深度检查，然后捕获所有词典并删除密码密钥：

from copy import deepcopy

def remove_pass(v):
    if isinstance(v, dict):
        if "password" in v:
            del v["password"]
        for ele in v.values():
            remove_pass(ele)
    elif isinstance(v, Iterable) and not isinstance(v, basestring):
        for ele in v:
            remove_pass(ele)


from pprint import pprint as pp
d = deepcopy(d)
for v in d.values():
   remove_pass(v)

输入：

{'_enabled': 'true',
 'foo': {'isdefault': 'false',
         'name': 'qa',
         'nested': {'password': 'nested'},
         'password': 'dasddf',
         'url': 'http://prod.example.com',
         'user': 'willow'},
 'instances': [{'isdefault': 'true',
                'name': 'dev',
                'password': 'abc123',
                'url': 'http://dev.example.com',
                'user': 'buffy'},
               {'isdefault': 'false',
                'name': 'prod',
                nested': {'more_nesting': {'even_more_nesting': ({'password': 'foobar'},
                                       {'password': 'foob'}),
                                        'password': 'bar'},
                'password': 'xxxxx',
                'url': 'http://prod.example.com',
                'user': 'spike'},
               {'isdefault': 'false',
                'name': 'qa',
                'password': 'dasddf',
                'url': 'http://prod.example.com',
                'user': 'willow'}],
 'label': 'MyServers'}

输出：

{'_enabled': 'true',
 'foo': {'isdefault': 'false',
         'name': 'qa',
         'nested': {},
         'url': 'http://prod.example.com',
         'user': 'willow'},
 'instances': [{'isdefault': 'true',
                'name': 'dev',
                'url': 'http://dev.example.com',
                'user': 'buffy'},
               {'isdefault': 'false',
                'name': 'prod',
                'nested': {'more_nesting': {'even_more_nesting': ({}, {})}},
                'url': 'http://prod.example.com',
                'user': 'spike'},
               {'isdefault': 'false',
                'name': 'qa',
                'url': 'http://prod.example.com',
                'user': 'willow'}],
 'label': 'MyServers'}

Answer 2

如果您知道每个数据的结构（即，在数组/字典的深度处期望＆＃34;密码＆＃34;键），这将很简单。您只需循环查看列表项和词典即可找到＆＃34;密码＆＃34;键。

如果每个设置字典的结构确实无法预测，那么您必须将解决方案混合在一起。我在这种情况下所做的是将我的JSON转储到字符串，使用正则表达式删除/隔离我感兴趣的数据，然后将字符串加载回结构化的JSON。

这样的事情：

导入json，重新

raw_data = """{
  "_enabled": true,
  "instances": [
    {
      "isdefault": true,
      "name": "dev",
      "password": "abc123",
      "url": "http://dev.example.com",
      "user": "buffy"
    },
    {
      "isdefault": false,
      "name": "prod",
      "password": "xxxxx",
      "url": "http://prod.example.com",
      "user": "spike"
    },
    {
      "isdefault": false,
      "name": "qa",
      "password": "dasddf",
      "url": "http://prod.example.com",
      "user": "willow"
    }
  ],
  "label": "MyServers"
}"""

# I load and then dump my raw_data just to iron out any inconsistencies 
# in formatting before applying regex. i.e., inconsistent use of " instead of '
structured_data = json.loads(raw_data)
dumped_data = json.dumps(structured_data)

scrubbed = re.sub(r'"password": ".*?",', '', dumped_data)
structured_scrubbed = json.loads(scrubbed)

结果：

structured_scrubbed = {'_enabled': True,
 'instances': [{'isdefault': True,
                'name': 'dev',
                'url': 'http://dev.example.com',
                'user': 'buffy'},
               {'isdefault': False,
                'name': 'prod',
                'url': 'http://prod.example.com',
                'user': 'spike'},
               {'isdefault': False,
                'name': 'qa',
                'url': 'http://prod.example.com',
                'user': 'willow'}],
 'label': 'MyServers'}

Answer 3

我使用通用函数getPaths来查找嵌套字典中特定键的路径。您可以使用它来查找“密码”密钥的所有路径，然后进行更改或删除。这适用于json字符串的所有格式/模式。

def getPaths(dictionary, searchKey):
    '''
    generator to get all paths for the key in the nested dictionary

    '''
    for k, v in dictionary.items():
        if k == searchKey :
            yield []
        elif isinstance(v, dict):
            # if the value if dict, go in recursively and yield the path
            for subkey in getPaths(v, searchKey):
                yield [k]+subkey
        elif isinstance(v, list):
            # if value is a list, for each element in the list, go in recursively and yield the path
            for i, item in enumerate(v):
                if isinstance(item, dict):
                    for subkey in getPaths(item, searchKey):
                        yield [k]+[i]+subkey


jsonstring = """{
  "_enabled": true,
  "instances": [
    {
      "isdefault": true,
      "name": "dev",
      "password": "abc123",
      "url": "http://dev.example.com",
      "user": "buffy"
    },
    {
      "isdefault": false,
      "name": "prod",
      "password": "xxxxx",
      "url": "http://prod.example.com",
      "user": "spike"
    },
    {
        "instance2": {
                      "isdefault": false,
                      "name": "qa",
                      "password": "dasddf",
                      "url": "http://prod.example.com",
                      "user": "willow"
                      }
    }
  ],
  "label": "MyServers"
}"""
import json
jsonObj = json.loads(jsonstring)


paths = getPaths(jsonObj , "password")
for path in paths:
    print('path:', path)

结果：

>>> path: ['instances', 0]
>>> path: ['instances', 1]
>>> path: ['instances', 2, 'instance2']

Answer 4

假设您只想检查列表或词典的容器，并从具有key = "password"

的词典条目中删除键值对

#first copy the structure
new_data = copy.deepcopy(data)

#this is a recursive function. 
#Heavily nested structures may fail due to recursion limit
def clean_hierarchy(ele):
    #lists may contain dictionaries, so clean theses entries
    if isinstance(ele,list):
        for val in ele:
            clean_hierarchy(val)
    if isinstance(ele,dict):
        #remove possible password entry
        if "password" in ele:
            ele.pop("password",None)
        #dictionary may contain more dictionaries. Rinse and repeat!
        for val in ele.values():
            clean_hierarchy(val)

clean_hierarchy(new_data)

如何擦除特定键的动态python字典？

4 个答案: